Answer:
7614
Explanation:
The exponential growth of the bacteria culture can be modeled by the equation ...
p = a·b^(t/c)
where 'a' is the initial population, 'b' is the growth factor, and 'c' is the period over which that growth factor applies.
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model
In the given scenario, we have ...
- a = 6000 bacteria
- b = 6600/6000 = 1.1 . . . . growth factor
- c = 4 hours
So, our model is ...
p = 6000·1.1^(t/4)
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application
After 10 hours, the population is predicted to be ...
p = 6000·1.1^(10/4) ≈ 7614
We predict 7614 bacteria will be present after 10 hours.
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Additional comment
If we compare the form we used to the one suggested in the proglem statement, we see
- p = 6000·1.1^(t/4)
- p = 6000·e^(kt)
For the bases of the exponential term to be the same, we must have ...
1.1^(t/4) = e^(kt)
(1.1^(1/4))^t = (e^k)^t . . . . . factor t from the exponent
1.1^(1/4) = e^k . . . . . . . match the bases
k = ln(1.1)/4 . . . . . take natural logs
k ≈ 0.0238275
and the equation becomes ...
p = 6000·e^(0.0238275t)
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In terms of 'a', 'b', and 'c' used above, we find ...
k = ln(b)/c